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If dy/dt = ky and k is a nonzero constant, y could be 
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#1
May210, 12:48 AM

P: 34

1. The problem statement, all variables and given/known data
If dy/dt = ky and k is a nonzero constant, than y could be a. 2e^kty b. 2e^kt c. e^kt + 3 d. kty + 5 e. .5ky^2 + .5 Correct answer is b. 2e^kt 2. Relevant equations 3. The attempt at a solution I don't really know what I should be doing. I have a feeling I have to work backwards, therefore do the integral of ky? But if that is the case, I have no idea how e^kt is created through that integral. 2e^kt reminded me of Ce^kt, but that doesn't have anything to do with this problem either. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
May210, 12:54 AM

P: 58

Look at the derivative of e^kt and see how that relates to e^kt.



#3
May210, 12:56 AM

HW Helper
P: 6,202

The 'correct' thing to do would be
dy/dt = ky ∫ dy/y = ∫ k dt But since you have a multiple choice, you could quickly differentiate each one if you wanted to. 


#4
May210, 01:07 AM

P: 34

If dy/dt = ky and k is a nonzero constant, y could be
Hm.. this is what I got when I did ∫ dy/y = ∫ k dt
lny = kt y = e^kt Where does the 2 come from? I guessed it would have to be from a square, but when you do the integral of a number (since k is a nonzero constant) you multiply the constant with a variable with an exponent of 1, not 2. 


#5
May210, 01:12 AM

HW Helper
P: 6,202

ln y = kt+A such that y=e^{kt+A}=Be^{kt} where B is a constant. So B can be any number, such as '2'. 


#6
May210, 01:14 AM

P: 34

Ah, I forgot about that. Thanks!



#7
May210, 08:16 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,564

dy/dt(1 2kt e^(kty)= 2k^y e^(kty). No, that differential equation is not satisfied. (I am assuming the parentheses combine these as you wanted that you did not intend 2(e^k)t.) 


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